Is ${152676}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {152676}= &&{1}\cdot100000+ \\&&{5}\cdot10000+ \\&&{2}\cdot1000+ \\&&{6}\cdot100+ \\&&{7}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {152676}= &&{1}(99999+1)+ \\&&{5}(9999+1)+ \\&&{2}(999+1)+ \\&&{6}(99+1)+ \\&&{7}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {152676}= &&\gray{1\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {1}+{5}+{2}+{6}+{7}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${152676}$ is divisible by $3$ if ${ 1}+{5}+{2}+{6}+{7}+{6}$ is divisible by $3$ Add the digits of ${152676}$ $ {1}+{5}+{2}+{6}+{7}+{6} = {27} $ If ${27}$ is divisible by $3$ , then ${152676}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${152676}$ must also be divisible by $3$.